Transcript Chapter 11 Electromagnetic Waves

Physics Beyond 2000
Chapter 11
Electromagnetic Waves
What are electromagnetic waves?
• EM waves are energy emitted resulting
from acceleration of electric charges.
-
EM Waves
•
•
•
•
They can travel through vacuum.
In vacuum, their speed = 3 × 108 ms-1
c = f.λ
An EM wave consists of electric and
magnetic fields, oscillating in phase and at
right angles to each other.
http://www.geo.mtu.edu/rs/back/spectrum/
Electromagnetic spectrum
• The range of the wavelength of EM waves
is enormous.
• 10-14 m – 1 km
• The electromagnetic spectrum is named
according to the range of the wavelength
and the method of production.
Radio Waves
• Production:
• Apply an a.c. voltage of high
frequency to a pair of metal rods
(dipole).
• If the rods are vertical, the radio wave
is also said to be vertically polarized.
transmitter
oscillating electric field
oscillating a.c.
direction of propagation of
radio wave
Radio Waves
• Receiver:
• The receiver dipole is parallel to the
direction of polarization. In this case, it
is vertical.
transmitter
receiver
oscillating electric field
oscillating a.c.
direction of propagation of
radio wave
Spectrum of Radio Waves
Radio waves
Wavelength
Long waves
1 km – 10 km
Medium waves
100 m – 1 km
Short waves
10 m – 100 m
Very high frequency
(VHF)
Ultra high frequency
(UHF)
1 m – 10 m
0.1 m – 1 m
Microwaves
• Microwave is polarized along the length of
the dipole.
transmitter
receiver
oscillating electric field
oscillating a.c.
direction of propagation of
microwave
Microwaves
• Vertical metal rod can absorb the energy of
the microwave.
metal rod
transmitter
receiver
oscillating electric field
oscillating a.c.
direction of propagation of
microwave
no response
Microwaves
• Horizontal metal rod cannot absorb the
energy of the microwave.
transmitter
metal rod
oscillating electric field
oscillating a.c.
direction of propagation of
microwave
receiver
Interference of Microwaves
• At P, the wave from the transmitter meets
the reflected wave. Interference occurs.
image
of transmitter
metal plate
P
transmitter
Interference of Microwaves
• We may consider it as an interference from two
coherent sources, the transmitter and its image.
image
of transmitter
metal plate
P
transmitter
Interference of Microwaves
• The two sources are in anti-phase because
there is a phase change of  on reflection.
image
of transmitter
metal plate
P
transmitter
Interference of Microwaves
• If the path difference at P = n., there is
destructive interference.
image
of transmitter
metal plate
P
transmitter
Interference of Microwaves
1
• If the path difference at P = (n  ), there
2
is constructive interference.
image
of transmitter
metal plate
P
transmitter
Microwave Cooking
• One possible frequency of microwave is
2.45 GHz which is equal to the natural
frequency of water molecules.
• Microwave can set water molecules into
oscillation. The water molecules absorb the
energy from microwave.
http://www.gallawa.com/microtech/howcook.html
Microwave in
satellite communications
• Reading the following
http://www.s-t.au.ac.th/~supoet/satel.htm#1
Infrared Radiation
• Self-reading.
Ultraviolet Radiation
• Self-reading.
Visible light
• Self-reading.
Colored Video Pictures
• Self-reading.
Scattering of Light
• Light energy is absorbed by an atom or
molecule.
• The atom (molecule) re-emits the light
energy in all direction.
• The intensity of light in initial direction is
reduced.
incident light
oscillating E-field
atom
Scattering of Light
• Light energy is absorbed by an atom or
molecule.
• The atom (molecule) re-emits the light
energy in all directions.
• The intensity of light in initial direction is
reduced.
axis along which the atom oscillating
scattered light
atom
Scattering of Light
• Note that there is not any scattered light
along the direction of oscillation of the atom.
• The scattered light is maximum at right
angle to the axis.
axis along which the atom oscillating
scattered light
strongest
strongest
atom
Why is the sky blue at noon and
red at sunrise and sunset?
• Why is the sky blue in daytime?
• http://physics.about.com/science/physics/lib
rary/weekly/aa051600a.htm
• Why is the sky red in sunset/sunrise?
• http://physics.about.com/science/physics/lib
rary/weekly/aa052300a.htm
Why is the sky blue at noon and
red at sunrise and sunset?
• At noon, we see the most scattered light.
• Note that the natural frequency of air molecules is
in the ultraviolet region. Blue light is easily
scattered by air molecules.
Blue light
is most scattered
white light from
the sun
Red light is least scattered
Why is the sky blue at noon and
red at sunrise and sunset?
• At sunset, we see the least scattered light.
• Red light is least scattered.
Red light
is least scattered
white light from
the sun
Blue light is most scattered
Polarization of light
• Light is transverse wave so it exhibits
polarization.
• Unpolarized light: the electric field is not
confined to oscillate in a plane.
• Plane-polarized light: the electric field at
every point oscillates in the same fixed
plane.
• Plane of polarization: the plane in which the
electric field of a plane polarized light
oscillates.
Polarization of Light
• Plane polarized light:
electric vector
• Unpolarised light:
electric vector
Polarization by Absorption
• An array of parallel conducting wires.
• It can absorb electric field of microwave
oscillating in a plane parallel to its conducting
wires.
Plane-polarized
microwave
Conducting wires are vertical
No microwave
E-field is vertial.
Polarization by Absorption
• An array of parallel conducting wires.
• It cannot absorb electric field of microwave
oscillating in a plane perdpndicular to its
conducting wires.
Plane-polarized
microwave
E-field is horizontal
Conducting wires are vertical
Plane-polarized
microwave
Polarization by Absorption
• An array of parallel conducting wires.
• It can be a polarizer of microwaves
Unpolarized
microwave
Conducting wires are vertical
Plane-polarized
microwave
Polarization by Absorption
• Polaroid is a plastic sheet consisting of long chains
of molecules parallel to one another.
• It can absorb electric field of light oscillating in a
plane parallel to its chains of molecules.
Plane-polarized light
Chains of molecules are vertical
No light
E-field is vertical
Polarization by Absorption
• Polaroid is a plastic sheet consisting of long chains
of molecules parallel to one another.
• It cannot absorb electric field of light oscillating in
a plane perpendicular to its chains of molecules.
Chains of molecules are vertical
Plane-polarized light
E-field is horizontal
Plane-polarized light
Polarization by Absorption
• Polaroid is a plastic sheet consisting of long chains
of molecules parallel to one another.
• It can be a polarizer of light.
Chains of molecules are vertical
Unpolarized light
Plane-polarized light
Polarization by Reflection
Assume that the direction of the reflected light and that of the refracted
light are perpendicular.
plane-polarized
incident light
No reflected light
air
glass
plane-polarized
refracted light
Polarization by Reflection
•The electric field sets the
electrons in the glass to
plane-polarized
oscillate at right angles to incident light
No reflected light
the refracted ray.
air
•The intensity perpendicular
glass
to the axis of oscillation is
plane-polarized
strongest  The refracted ray
refracted light
is bright.
•The intensity parallel to
the axis of oscillation is zero
 no reflected ray.
Polarization by Reflection
Assume that the direction of the reflected light and that of the refracted
light are perpendicular.
The plane of polarization
Unpolarized
is parallel to the surface
incident light
of medium.
Polarized reflected light
air
glass
Unpolarized
refracted light
The Brewster’s Angle
• p = Brewster’s angle
• r = Angle of refraction.
incident
ray
p p
reflected
ray is completely
polarized
air
medium
r
refracted
ray
The Brewster’s Angle
• n = tan p where n is the refractive index of
the medium.
incident
ray
p p
reflected
ray
air
medium
Prove it!
r
refracted
ray
Example 1
• The Brewster’s angle for glass is about
56.3o.
Polarization by Scattering
• When light energy is absorbed by an atom,
the atom re-radiates the light.
incident ray
atom
The atom absorbs
the wave energy.
The atom re-radiates
the wave energy.
Polarization by Scattering
no scattered light
vertically polarized
light
vertically polarized
light
water mixed with milk
vertically polarized
light
vertically polarized
light
no scattered light
Polarization by Scattering
horizontally polarized
light
no scattered light
horizontally polarized
light
water mixed with milk
horizontally polarized
light
no scattered light
horizontally polarized
light
Polarization by Scattering
horizontally polarized
light
unpolarized
light
vertically polarized
light
water mixed with milk
unpolarized
light
vertically polarized
light
horizontally polarized
light
Polaroid Sunglasses
• Why are the polaroid sunglasses designed to
absorb horizontally polarized light?
Study p.233 of the textbook.
Interference of Light
• Light is a kind of wave.
• Interference is a wave property.
Interference of Light
Conditions for an observable interference pattern of
light:
• Coherent sources : two sources emit light of
the same frequency and maintain a constant
phase difference.
• The light waves are of same frequency and
almost equal amplitude.
• The separation of the two sources is of the
same order as the wavelength.
• The path difference must be not too large.
Interference of Light
• Young’s double-slit experiment
http://surendranath.tripod.com/DblSlt/DblSltApp.html
http://members.tripod.com/~vsg/interfer.htm
•The incident ray is split into two coherent sources
S1 and S2 by the double-slit.
•S1 and S2 are in phase.
•The screen is far away from the slit. D>> a.
•The angles are very small.
Young’s double-slit experiment
Suppose that there is a maximum at point P.
A constructive interference occurs at P.
S1
P

a
central line
S2
D
screen
Young’s double-slit experiment
As point P is far away from the double slit,
the light rays of the same fringe are parallel.
parallel rays meet at point P

S1
a

S2 
central line
The path difference  = a.sin
Young’s double-slit experiment
For points with constructive interference,
= a.sin = m. where m = 0, 1, 2,…
m is the order of the fringes.

S1
a

S2 
parallel rays meet at point P with
maximum intensity.
central line
The path difference  = a.sin
Young’s double-slit experiment
For points with destructive interference,
1
= a.sin = (m + ).  where m = 0, 1, 2,…
2

S1
a

S2 
parallel rays meet at a point with
minimumintensity.
central line
The path difference  = a.sin
separation between the fringes
Suppose the the order of the fringe at P is m.
The distance from P to the central line is ym.
The distance between the double slit
and the screen is D.
P
ym

S1
a M
S2

central line
D
separation between the fringes
The line from mid-point M to P makes the
same angle  with the central line.
ym = D.tan  D.sin  =
P
m
.D
a
ym

S1
a M
S2

central line
D
separation between the fringes
ym =
m
.D
a
(1)
By similar consideration, for the m+1 bright fringe
ym+1=
(m  1)
.D
a
The separation between the two fringes is
s = ym+1 – ym =
D
a
(2)
separation between the fringes
s = ym+1 – ym =
D
a
• The fringes are evenly separated.
• For well separated fringes.
• s  D  Place the screen far away from the slit.
• s   Different separation for waves of different
wavelength.
• s  1  The slits should be close.
a
Variation of intensity
• If the slits are sufficiently narrow, light spreads out
evenly from each slit and the bright fringes are
equally bright.
Variation of intensity
• If the intensity on the screen using one slit
is Io,
the intensity is 4.Io at the position of bright
fringes and
the intensity is 0 at the position of dark
fringes.
• Energy is re-distributed on the screen.
Variation of intensity
• In practice, light waves do not diffract
evenly out from each slit. There is an angle
of spread.
http://numerix.us.es/numex/numex2.html
http://bc1.lbl.gov/CBP_pages/educational/java/duality/duality2.html
Variation of intensity
• The intensity of the fringes is enclosed in an
envelope as shown.
Example 2
• Separation of fringes in Young’s double-slit
experiment.
White light fringes
s=
D
a
 s
Separation of violet fringes is shortest.
Separation of red fringes is longest.
http://surendranath.tripod.com/DblSlt/DblSltApp.html
http://members.tripod.com/~vsg/interfer.htm
Submerging in a Liquid
• If the Young’s double-slit experiment is
done in a liquid, what would happen to the
separation of fringes?
ym
Liquid with refractive index n
Submerging in a Liquid
• The wavelength changes!
Let  be the wavelength in vacuum/air
and n the wavelength in liquid.
Let n be the refractive index of the liquid.
n
sn =
s
n
n 

The fringe separation in liquid
Submerging in a Liquid
• The fringe separation is reduced by a factor
of n.
n
sn =
s
n
n 

The fringe separation in liquid
Optical path
• Optical path of light in a medium is the
equivalent distance travelled by light in
vacuum.
thickness = t
medium of
refractive index
n
Incident
light
Light requires the same
time to travel through
thickness = optical path
the two paths.
vacuum
Optical path
• Show that the optical path = n.t
thickness = t
medium of
refractive index
n
Incident
light
Light requires the same
time to travel through
thickness = optical path
the two paths.
vacuum
Optical path
• The number of waves in the medium = the
number of waves in the optical path
thickness = t
medium of
refractive index
n
Incident
light
Light requires the same
time to travel through
thickness = optical path
the two paths.
vacuum
Example 3
• Note that light rays pass through different
media. We need to consider their path
difference in terms of the optical paths.
Shifting a System of Fringes
Note that ray A has passes through a medium of refractive
index n and thickness t.
We need to find the path difference in terms of the optical
path.
A
B
Shifting a System of Fringes
Without the medium, the central maximum is at
the central line. (Path difference = 0)
Now the central maximum shifts to another position.
Find the central maximum.
The central maximum shifts through a distance
(n  1)tD
y
a
Shifting a System of Fringes
If the central line now has the mth bright fringe,
the central maximum has shifted through m fringes.
m
( n  1)t

Multiple-slit
• More slits than two.
3 slits
4 slits
http://bednorzmuller87.phys.cmu.edu/demonstrations/optics/interference/demo323.html
Multiple-slit
• N = number of slits.
• Compare N = 2 with N = 3
http://wug.physics.uiuc.edu/courses/phys114/spring01/Discussions/html/wk3/multiple/html/3-extra2.htm
Multiple-slit
• N = number of slits.
• Compare N = 2 with N = 3
N=2
N=3
Maximum occurs at positions
with a.sin = m.
Maximum occurs at positions
with a.sin = m.
The intensity at the maximum is
4.Io
The intensity at the maximum is
9.Io
Between two maxima, it is a
minimum.
Between two maxima, it is a peak.
The width of bright fringes is
large.
The width of bright fringes is less
Multiple-slit
• N = number of slits.
• Compare N = 2 with large N.
N=2
Large N
Maximum occurs at positions
with a.sin = m.
Maximum occurs at positions
with a.sin = m.
The intensity at the maximum is
4.Io
The intensity at the maximum is
N2.Io
Between two maxima, it is a
minimum.
Between two maxima, there are
(N-2) peaks. The intensity of the
peak is almost zero
The width of bright fringes is
very narrow
Multiple-slit
• N=3
• Maximum occurs at a.sin = m. (same as N = 2).
Central maximum
1st order maximum
phase difference
=0
a
a
a
a
m= 0   = 0
All three rays are in phase.
phase difference
=0
m= 1  a.sin = 
All three rays are in phase.
Multiple-slit
• Textbook, p.238. Fig. 30.
• Position of maximum when N = 2 is still a
maximum.
• There are peak(s) between two successive
maxima.
• The number of peaks = N – 2.
• The intensity of peaks drops with N.
Multiple-slit (N = 3)
m=1
m=0
m=1
 sin   
sin  
2a
2a
 sin   0 sin   
sin  
a
a
Why is there a peak between two maxima when N = 3?
Multiple-slit (N = 3)
There is a peak at position with sin  
Δ1 =
a
θ

2

2a
phase difference
=π
θ
Δ1
a
Δ2 =λ phase difference
=0
Δ2
Add three rotating vectors for the resultant wave.
Multiple-slit (N = 3)
m=1
m=0
m=1
 sin   
sin  
2a
2a
sin  
2
3a
sin  

sin   0
sin  

a
3a
Why are there 2 minima between two maxima when N = 3?
Multiple-slit (N = 3)
There is a minimum at position with sin  
Δ1 =
a

3
θ
Δ1
a
2
3
2
Δ2 =
phase difference
3
=
Δ2
3a
phase difference
=
θ

4
3
Add three rotating vectors for the resultant wave.
Multiple-slit (N = 3)
2
There is a minimum at position with sin  
3a
2
Δ1 =
phase difference
3
a
=
θ
θ
Δ1
a
4
Δ2 =
phase difference
3
=
Δ2
4
3
8
3
Add three rotating vectors for the resultant wave.
Diffraction grating
• A diffraction grating is a piece of glass with
many equidistant parallel lines.
Diffraction grating
• The mth order maximum is given by
m=3
a.sin = m.
m=2
m=1
m=0
m=1
m=2
m=3
Diffraction grating
• The maximum order is given by
mmax 
a

m=3
m=2
m=1
m=0
m=1
m=2
m=3
Intensity of diffraction grating
coarse grating and fine grating
• The separation between lines on a coarse
grating is longer than that of a fine grating.
• Example of a coarse grating: 300 lines/cm.
• Example of a fine grating: 3000 lines/cm.
Find the maximum order of the above two gratings.
Colour Spectrum from a
Diffraction Grating
a..sinm = m. 
The spatial angle m depends on 
m=3
m=2
m=3
m=2
m=1
m=1
white light
m=0
m=1
m=1
m=2
m=3
m=3
m=2
Example 4
• Monochromatic light = light with only one
colour (frequency)
Example 5
• Overlapping of colour spectrum
Blooming of lenses
• Coat a thin film on a lens to reduce the
reflection of light.
reflected ray
incident ray
incident ray
no reflected ray
film
glass
glass
without coating
with coating
Blooming of lenses
• Ray A is reflected at the boundary between
air and the film.
Reflected ray A
incident ray
with coating
Blooming of lenses
• Ray B is reflected at the boundary between
the thin film and the glass.
Reflected ray A
incident ray
Reflected ray B
with coating
Blooming of lenses
• It is designed to have destructive interference for
the reflected light rays.  No reflected light ray.
Reflected ray A
incident ray
Reflected ray B
with coating
Blooming of lenses
• Suppose that the incident ray is normal to
the lens.
• The reflected light rays are also along the
normal.
reflected rays
incident ray
film
film
glass
glass
Blooming of lenses
• Let n’ be the refractive index and t be the
thickness of the thin film.
• Let  be the wavelength of the incident light.
incident ray
reflected rays
film
film
glass
glass
Blooming of lenses
• To have destructive interference for the
reflected rays,

t min 
4n'
incident ray
reflected rays
film
film
glass
glass
Blooming of lenses
• Energy is conserved. As there is not any
reflected light rays, the energy goes to the
transmitted light ray.
incident ray
No reflected rays
film
film
glass
glass
Blooming of lenses
• The reflected ray from the bottom of the
glass is so dim that it can be ignored.
This reflected ray
is ignored.
incident ray
film
film
glass
glass
Blooming of lenses
• Limitation:
• For normal incident ray only.
• For wave of one particular wavelength only.
This reflected ray
is ignored.
incident ray
film
film
glass
glass
Examples
• Example 6
Find the minimum thickness of the thin
film.
• Example 7
Find the wavelength.
Air Wedge:
Experimental setup
Air Wedge
A normal incident ray is reflected at the boundary
between the slide and the air wedge.
reflected ray A
normal incident
ray
slide
air wedge
glass block
Air Wedge
The normal incident ray goes into the wedge, passing
through a distance t and reflected at the boundary between
the air wedge and the glass block.
reflected ray A
normal incident
ray
reflected ray B
slide
t
air wedge
glass block
Air Wedge
The ray into the glass block is ignored.
reflected ray A
normal incident
ray
reflected ray B
slide
t
air wedge
glass block
Air Wedge
The ray reflected at the top of the slide is ignored.
normal incident
ray
slide
air wedge
glass block
Air Wedge
• Depending on the the distance t and the
wavelength  of the incident wave, the two
reflected rays may have interference.
• The pattern is a series of bright and dark
fringes when we view through the travelling
microscope.
Air Wedge
1
To produce bright fringes, 2.t = (m - )., m = 1, 2, 3,…
2
reflected ray A
normal incident
ray
reflected ray B
constructive
interference
slide
t
air wedge
glass block
Air Wedge
•Ray B has a phase change  on reflection.
•The path difference must be m.
reflected ray A
normal incident
ray
reflected ray B
constructive
interference
slide
t
air wedge
glass block
Air Wedge
To produce dark fringes, 2.t = m ., m = 0, 1, 2,…
reflected ray A
normal incident
ray
reflected ray B
destructive
interference
slide
t
air wedge
Note that ray B has a phase change on reflection.
glass block
Air Wedge
•Ray B has a phase change  on reflection.
1
•The path difference must be (m + ).
2
reflected ray A
normal incident
ray
reflected ray B
destructive
interference
slide
t
air wedge
Note that ray B has a phase change on reflection.
glass block
Air Wedge
• At the vertex, t = 0  m = 0  dark fringe.
reflected ray A
normal incident
ray
reflected ray B
destructive
interference
slide
t
air wedge
Note that ray B has a phase change on reflection.
glass block
Air Wedge
To find the separation s between two successive
bright fringes
Let D be the height of the high end of the slide.
Let L be the length of the slide.
Nth fringe
(N+1)th fringe
L
 tN s
tN+1
slide
D
air wedge
glass block
Air Wedge
To find the separation s between two successive
bright fringes Angle of inclination of the slide can be found from
D
sin  
L
Nth fringe
(N+1)th fringe
L
 tN s
tN+1
slide
D
air wedge
glass block
Air Wedge
Separation s between two successive bright fringes

tN+1 – tN =
2


2
 tan  

s
2.s

tN+1

tN
s
Air Wedge
Separation s between two successive bright fringes
and
D
sin  
L
For small angle , sin   tan 
tan  


2.s
L.
s
2 .D

tN+1

tN
s
Air Wedge
• For flat surfaces of glass block and slide
– the fringes are parallel and evenly spaced.
Air Wedge
• For flat surface of glass block and slide with
surface curved upwards
– the fringes are parallel and become more
closely packed at higher orders.
Air Wedge
• For flat surface of glass block and slide with
surface curved downwards
– the fringes are parallel and become more
widely separated at higher orders.
Air Wedge
• We can use this method to check a flat glass
surface.
The surface is flat.
The surface is not flat.
Measuring the Diameter D
of a Wire
L.
D
2.s
Measure the quantities on the right
hand side and calculate D.
L
air wedge
D
Example 8
• There are 20 dark fringes  m = 19.
L
19.
• L  19.s  s =
D= 2
19
m = 19
m=0
L
19.s
D
Soap film
• Why soap film is coloured?
http://www.cs.utah.edu/~zhukov/applets/film/applet.html
Soap film
The two transmitted rays A and B may have interference
depending on the thickness t of the film and the wavelength
.
t
ray A
incident ray
ray B
soap
water
Soap film
Note that ray B has two reflections. No phase change is due
to reflection.
t
ray A
incident ray
ray B
Soap film
To observe bright fringes, the path difference = m.
 2.t = m. 
t
ray A
constructive
interference
incident ray
ray B
Soap film
1
To observe dark fringes, the path difference = (m+ ).
2
 2.t = (m+ 1 ). 
2
t
ray A
constructive
interference
incident ray
ray B
Soap film
The two reflected rays A and B may have interference
depending on the thickness t of the film and the wavelength
.
incident ray
t
ray A
ray B
soap
water
Soap film
Note that this time there is a phase change due to reflection.
incident ray
t
ray A
ray B
soap
water
Soap film
Find out how the interference depends on t and 
incident ray
t
ray A
interference
ray B
soap
water
Soap film
• As the interference depends on , there will
be a colour band for white incident light.
• In each colour band, violet is at the top and
red is at the bottom.
Soap film
• Soap water tends to move downwards due
to gravity.
• The soap film has a thin vertex and a thick
base. The fringes are not evenly spaced.
• The fringes are dense near the bottom and
less dense near the vertex.
Soap film
• The pattern of the reflected rays and that of
the transmitted rays are complementary.
incident ray
C: constructive
interference
D: destructive
interference
D
C
D
C
reflected rays
C
D
C
D
transmitted rays
Example 9
• The reflected rays have constructive
interference.
• Note that there is a phase change on
reflection.
Newton’s rings:
Experimental setup
Newton’s rings
• What do we see through the travelling
microscope with white incident light?
Newton’s rings
• If we use red incident light,
http://www.cs.utah.edu/~zhukov/applets/film/applet.html
Newton’s rings
The two reflected rays have interference depending on
the thickness t of the air gap and the wavelength .
reflected ray A
reflected ray B
incident ray
interference
lens
air
t
glass block
t = thickness of air gap
Newton’s rings
For bright fringes, t 
(2m  1).
4
reflected ray A
reflected ray B
incident ray
interference
lens
air
t
glass block
t = thickness of air gap
Newton’s rings
m.
For dark fringes, t  2
reflected ray A
reflected ray B
incident ray
interference
lens
air
t
glass block
t = thickness of air gap
Newton’s rings
At the center of the lens, there is a dark spot.
reflected ray A
reflected ray B
incident ray
interference
lens
air
t
glass block
t = thickness of air gap
Newton’s rings
• The spacing of the rings are not even.
• Near the center, the rings are widely
separated.
• Near the edge, the rings are close together.
Newton’s rings
• Find the radius of the mth dark ring.
Rm
Newton’s rings
C
Rm  R  ( R  t )  2tR
2
2
R = radius of curvature of the lens
lens
A
Rm
t
O
glass block
B
t
air
t = thickness of air gap
Newton’s rings
C
Rm  2tR
m for the mth dark fringe
t
2
R = radius of curvature of the lens
lens
A
Rm
t
O
glass block
B
t
air
t = thickness of air gap
Newton’s rings
C
Rm  mR
R = radius of curvature of the lens
lens
A
Rm
t
O
glass block
B
t
air
t = thickness of air gap
Newton’s rings
• Separation between two successive rings
s = Rm+1 – Rm = ( m  1  m ). R
The separation approaches zero for high orders.
Example 10
• To find the radius of curvature of a lens by
Newton’s rings.
• If there is distortion of the Newton’s rings,
the lens is not a good one.
Thin films
Light is incident obliquely onto a thin film of refractive
index n and thickness t.
incident
ray
n


t
thin film
Thin films
The reflected rays have interference depending on the angle
of view  and wavelength .
incident
ray
n


t
thin film
Thin film
incident white light
coloured
spectrum
http://www.cs.utah.edu/~zhukov/applets/film/applet.html
Diffraction of Light
laser tube
single slit
screen
http://surendranath.tripod.com/SnglSlt/SnglSltApp.html
Diffraction of Light
• Intensity variation
http://arborsci.com/Oscillations_Waves/Diffraction.htm
Diffraction of Light
• Consider a light source which is far away
from the single slit and the light is normal to
the slit. http://www-optics.unine.ch/research/microoptics/RigDiffraction/aper/aper.html
• In general, the position of the mth dark
fringe due to a slit of width d is given by
d.sinm = m.
Diffraction of Light
• Variation of intensity
• http://www.fed.cuhk.edu.hk/sci_lab/downlo
ad/project/javapm/java/slitdiffr/Default.htm
Theory of diffraction
Formation of 1st order dark fringe.
Divide the slit into two equal sections A1 and A2.
The light from section A1 cancels the light from section A2
Prove that
A11
1
d
A2
sin 1 

d
Theory of diffraction
Formation of 2nd order dark fringe.
Divide the slit into 4 equal sections A1 , A2, A3 and A4.
The light from section A1 cancels the light from section A2.
The light from section A3 cancels the light from section A4
A1
d
A2
A3
A4
2
2
Prove that sin  2 
d
Theory of diffraction
Formation of 2nd order dark fringe.
In general for the mth dark fringe,
A1
d
A2
A3
A4
2
m
sin  m 
d
Diffraction of Light
• Consider a light source which is far away
from the single slit and the light is normal
to the slit.
• For a circular hole with diameter d, the
center is a bright spot and the 1st dark ring
is given by
1.22
sin 1 
d